Question

Suppose that v1(t)=100cos(ωt)and v2(t)=100sin(ωt). Use phasors to reduce the sum vs(t)=v1(t)+v2(t) to a single term of the form Vmcos(ωt+θ). Enter your answers numerically separated by a comma. Express θ in degrees using three significant figures

Answer 1

Answer:

$vs(t)=152.08cos(\omega t -45)=152.08\angle-45$

Step-by-step explanation:

A tension expressed in form:

$V(t)=Vm*cos(\omega t+\phi)$

It can be converted directly into a fasor:

$V(t)=Vm\angle \phi$

If the function is a positive sine we subtract 90 °

$V(t)=Vm*sin(\omega t+\phi)=Vm*cos(\omega t+\phi-90)$

So:

$v1(t)=100cos(\omega t)= 100\angle0\\v2(t)=100sin(\omega t)=100cos(\omega t-\ph90)=100\angle -90$

$vs(t)=v1(t)+v2(t)=100\angle 0 +100\angle-90=152.0811931\angle-45\approx152.08\angle-45$

Therefore:

$vs(t)=152.08\angle-45=152.08cos(\omega t -45)$